Is there a better vectorization technique?

3 次查看（过去 30 天）
Kalamaya
Kalamaya 2012-8-29
I am trying to see if there are other ways of coding this code sample more efficiently. Here, y is an 1xM matrix, (say, 1x1000), and z is an NxM matrix, (say, 5x1000).
mean(ones(N,1)*y.^3 .* z,2)
This code works fine, but I worry of N increases a lot, that the:
ones(N,1)*y.^3
might get too wasteful and make everything slow down.
Thoughts?

请先登录，再回答此问题。

采纳的回答

John D'Errico
John D'Errico 2012-8-29
Its not THAT terrible for a matrix that small. However, you can gain from the use of bsxfun in many instances. Here, the matrices are simply too small to really gain anything.
>> N = 5;M =1000;
>> y = rand(1,M);
>> z = rand(N,M);
>> mean(ones(N,1)*y.^3 .* z,2)
ans =
0.12412
0.11669
0.12102
0.11976
0.12196
>> mean(bsxfun(@times,y.^3,z),2)
ans =
0.12412
0.11669
0.12102
0.11976
0.12196
>> z*y.'.^3/M
ans =
0.12412
0.11669
0.12102
0.11976
0.12196
As you can see, all three solutions return the same result. All are equally valid.
Now I'll compare the times required.
>> timeit(@() mean(ones(N,1)*y.^3 .* z,2))
ans =
0.00023018
>> timeit(@() mean(bsxfun(@times,y.^3,z),2))
ans =
0.00026829
>> timeit(@() z*y.'.^3/M)
ans =
0.00016594
As I said, you don't gain much. In fact, bsxfun does not gain at all, and is a bit slower. But you can gain a bit, if you re-write the expression into the third form I've posed. Not much, but a bit.
  2 个评论
Kalamaya
Kalamaya 2012-8-29
I thought about using bsxfun. How is the timing if the N was very large, say, 2000 or something like that? (timeit is your own function?)

请先登录，再进行评论。

更多回答（1 个）

Jan
Jan 2012-8-29
编辑：Jan 2012-8-30
The power() operator is more expensive than the matrix multiplication. Therefore an explicit multiplication saves time:
M = 1000;
N = 5;
y = rand(1, M);
z = rand(N, M);
tic; for i=1:100, a = mean(ones(N,1) * y .^ 3 .* z, 2); end; toc
% Elapsed time is 0.036668 seconds.
tic; for i=1:100, a = z * y.' .^ 3 / M; end; toc
% Elapsed time is 0.026818 seconds.
tic; for i=1:100, a = z * (y .* y .* y)' / M; end; toc
% Elapsed time is 0.001327 seconds.
[EDITED] If the resulting array is very large, a multiplication is faster than a division, but the result can differ due to rounding:
a = z * (y .* y .* y)' * (1 / M);
For the small [5x1] array in the example this does not matter.

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by